A uniform electric field E=(8m/e)V/m is created between two parallel plates of length 1 m as shown in figure, (where m= mass of electron and e= charge of electron). An electron enters the field symmetrically between the plates with a speed of 2m/s. The angle of the deviation ( θ ) of the path of the electron as it comes out of the field will be .........
(A)tan−1(4)
(B)tan−1 (2)✓ Correct
(C)tan−1(31)
(D)tan−1(3)
Answer: (B)
Solution
The acceleration in the y-direction is ay=eE/m=e(8m/e)/m=8 m/s². Time to traverse the plates: t=L/vx=1/2=0.5 s. Final y-velocity: vy=ayt=8×0.5=4 m/s. The deflection angle: tanθ=vy/vx=4/2=2, so θ=tan−1(2).
JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option